Divide the following rational expressions and simplify the result. $\dfrac{x^2+16xz+64z^2}{x^2z+8xz^2}\div\dfrac{x^2+5xz-24z^2}{x^4z-9x^2z^3}=$
Explanation: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $x^2+16xz+64z^2$, of the dividend can be factored as $(x+8z)(x+8z)$ using the perfect square pattern. The denominator, $x^2z+8xz^2$, of the dividend can be factored as $xz(x+8z)$ by factoring out $xz$. The numerator, $x^2+5xz-24z^2$, of the divisor can be factored as $(x+8z)(x-3z)$ using the sum-product pattern. The denominator, $x^4z-9x^2z^3$, of the divisor can be factored as $x^2z(x+3z)(x-3z)$ by factoring out $x^2z$ and using the difference of squares pattern. Now the division looks as follows: $\dfrac{(x+8z)(x+8z)}{xz(x+8z)}\div\dfrac{(x+8z)(x-3z)}{x^2z(x+3z)(x-3z)}$ To divide two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\phantom{=} \dfrac{(x+8z)(x+8z)}{xz(x+8z)}\div\dfrac{(x+8z)(x-3z)}{x^2z(x+3z)(x-3z)}$ $\begin{aligned} &\!= \dfrac{(x+8z)(x+8z)}{xz(x+8z)} \cdot \dfrac{x^2z(x+3z)(x-3z)}{(x+8z)(x-3z)} &\text{Flip the divisor.}\\\\\\\\ &\!= \dfrac{(x+8z)(x+8z) \cdot x^2z(x+3z)(x-3z)}{xz(x+8z) \cdot (x+8z)(x-3z)} &\text{Multiply across.}\\\\\\\\ &\!=\! \dfrac{{\cancel{(x\!+\!8z)}}\!{\cancel{(x\!+\!8z)}} \!{\cancel{x}}\!\cdot\!x{\cancel{z}}(x\!+\!3z)\!{\cancel{(x\!-\!3z)}}}{{\cancel{x}}{\cancel{z}}{\cancel{(x\!+\!8z)}}{\cancel{ (x\!+\!8z)}}{\cancel{(x\!-\!3z)}}} &\text{Cancel common factors.}\\\\\\\\ &\!=x(x+3z) \end{aligned}$ Therefore, the simplified form of the quotient is $x\left(x+3z\right)$, which is equivalent to $x^2+3xz$.